$\dfrac{ -7s - 6t }{ 3 } = \dfrac{ -s - 3u }{ -2 }$ Solve for $s$.
Answer: Multiply both sides by the left denominator. $\dfrac{ -7s - 6t }{ {3} } = \dfrac{ -s - 3u }{ -2 }$ ${3} \cdot \dfrac{ -7s - 6t }{ {3} } = {3} \cdot \dfrac{ -s - 3u }{ -2 }$ $-7s - 6t = {3} \cdot \dfrac { -s - 3u }{ -2 }$ Multiply both sides by the right denominator. $-7s - 6t = 3 \cdot \dfrac{ -s - 3u }{ -{2} }$ $-{2} \cdot \left( -7s - 6t \right) = -{2} \cdot 3 \cdot \dfrac{ -s - 3u }{ -{2} }$ $-{2} \cdot \left( -7s - 6t \right) = 3 \cdot \left( -s - 3u \right)$ Distribute both sides $-{2} \cdot \left( -7s - 6t \right) = {3} \cdot \left( -s - 3u \right)$ ${14}s + {12}t = -{3}s - {9}u$ Combine $s$ terms on the left. ${14s} + 12t = -{3s} - 9u$ ${17s} + 12t = -9u$ Move the $t$ term to the right. $17s + {12t} = -9u$ $17s = -9u - {12t}$ Isolate $s$ by dividing both sides by its coefficient. ${17}s = -9u - 12t$ $s = \dfrac{ -9u - 12t }{ {17} }$